A neutron with kinetic energy K = 10 MeV activates a nuclear reaction
n + 12 C → 9 Be + α
Find the kinetic energy of the α -particles outgoing at right angle to the direction of incoming neutrons. [Take u = 931.1 MeV and threshold energy of reaction (E th ) = 6.17 MeV]
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Sol. From conservation of energy, we have
m 1 c 2 + m 2 c 2 + K 1 = m 3 c 2 + m 4 c 2 + K 3 + K 4
or Q + K 1 = K 3 + K 4 … (1)
From momentum conservation along x-axis, we get 
= 
cos θ
or m 1 K 1 = m 4 K 4 cos 2 θ … (2)
and along y-axis, 
= 
sin θ
or m 3 K 3 = m 4 K 4 sin 2 θ … (3)
Adding eqn. (2) and eqn. (3),
m 1 K 1 + m 3 K 3 = m 4 K 4 or K 4 = 
K 1 + 
K 3
Substituting the value of K 4 is eqn. (1), we get
Q + 
K 1 = 
K 3
Here, Q = 
= 
= – 6.17 
= – 5.69
Thus 
K 3 = – 5.69 + 
(10)
K 3 = 
[ – 5.69 + 8.89] = 2.21 MeV
K 3 = 2.21 MeV
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